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Indefinite Integrals — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsIndefinite Integrals4 Marks
Question
Evaluate the following intregals:
$\int\frac{\text{x}+1}{\sqrt{4+5\text{x}-\text{x}}}\text{ dx}$
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Answer

Let $\text{I}=\int\frac{\text{x}+1}{\sqrt{4+5\text{x}-\text{x}^2}}\text{ dx}$
Let $\text{x}+1=\lambda\frac{\text{d}}{\text{dx}}(4+5\text{x}-\text{x}^2)+\mu$
$=\lambda(5-2\text{x})+\mu$
$\text{x}=(-2\lambda)\text{x}+5\lambda+\mu$
comparing the coefficient of like powers of x,
$-2\lambda=1\ \Rightarrow\lambda=-\frac{1}{2}$
$5\lambda+\mu=1\ \Rightarrow5\Big(-\frac{1}{2}\Big)+\mu=1$
$\mu=\frac{7}{2}$
So, $\text{I}=\int\frac{-\frac{1}{2}(5-2\text{x})+\frac{7}{2}}{\sqrt{4+\text{x}-\text{x}^2}}\ \text{dx}$
$=-\frac{1}{2}\int\frac{(5-2\text{x})}{\sqrt{4+5\text{x}-\text{x}^2}}\ \text{ dx}+\frac{7}{2}\int\frac{1}{\sqrt{-\big[\text{x}^2-2\text{x}\big(\frac{5}{2}\big)+\big(\frac{5}{2}\big)^2-\big(\frac{5}{2}\big)^2-4\big]}}\text{ dx}$
$\text{I}=-\frac{1}{2}\int\frac{5-2\text{x}}{\sqrt{4+5\text{x}-\text{x}^2}}\text{dx}+\frac{7}{2}\int\frac{1}{\sqrt{\Big[\Big(\frac{\sqrt{41}}{2}}\Big)^2-\Big(\text{x}-\frac52\Big)^2\Big]}\text{dx}$
$\text{I}=-\frac{1}{2}\int\frac{5-2\text{x}}{\sqrt{4+5\text{x}-\text{x}^2}}\text{dx}+\frac{7}{2}\int\frac{1}{\sqrt{-\Big[\big(\text{x}-\frac{\sqrt{41}}{2}\big)^2-\big(\text{x}-\frac{5}{2}^2}\Big]}\text{ dx}$
$\text{I}=\frac{1}{2}(2\sqrt{4+5\text{x}-\text{x}^2})+\frac{7}{2}\sin^{-1}\Bigg(\frac{\text{x}-\frac{5}{2}}{\frac{\sqrt{41}}{2}}\Bigg)+\text{c}$ $\big[\text{since},\int\frac{1}{\sqrt{\text{x}}}\text{dx}=2\sqrt{\text{x}}+\text{c},\int\frac{1}{\sqrt{\text{a}^2-\text{x}^2}}\text{dx}=\sin^{-1}\big(\frac{\text{x}}{\text{a}}\big)+\text{c}\big]$
$\text{I}=-\sqrt{4+5\text{x}-\text{x}^2}+\frac{7}{2}\sin^{-1}\Big(\frac{2\text{x}-5}{\sqrt{41}}\Big)+\text{c}$

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